Binary Tree Node Structure in DSA Javascript - Execution Trace

Concept Flow - Binary Tree Node Structure

Create new node

↓

Assign data value

↓

Set left child pointer to null

↓

Set right child pointer to null

↓

Node ready for use in tree

This flow shows how a binary tree node is created with data and two child pointers initialized to null.

Execution Sample

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

This code defines a binary tree node with data and two pointers for left and right children.

Execution Table

Step	Operation	Node Data	Left Pointer	Right Pointer	Visual State
1	Create node with data=10	10	null	null	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘
2	Node ready for use	10	null	null	┌────────┐ │ data:10│ │ left:∅ │ │right:∅ │ └────────┘

💡 Node creation complete with data=10 and both child pointers set to null.

Variable Tracker

Variable	Start	After Step 1	Final
node.data	undefined	10	10
node.left	undefined	null	null
node.right	undefined	null	null

Key Moments - 2 Insights

Why are left and right pointers set to null initially?

Because the node starts without children, setting left and right to null shows no child nodes exist yet, as seen in execution_table step 1.

What does the 'data' field represent in the node?

It holds the value stored in the node, which is 10 in the example, shown in execution_table step 1 under Node Data.

Visual Quiz - 3 Questions

Look at the execution_table at step 1, what is the value of node.left?

A 10

B undefined

C null

D node.right

💡 Hint

Check the 'Left Pointer' column in execution_table row for step 1.

At which step is the node fully created with data and pointers set?

A Step 1

B Step 2

C Step 3

D Step 0

💡 Hint

Look at the 'Operation' column and final visual state in execution_table.

If we add a left child node, which pointer changes from null?

A node.left

B node.right

C node.data

D node itself

💡 Hint

Recall that left and right pointers hold child nodes; left child changes node.left.

Concept Snapshot

Binary Tree Node Structure:
- Each node stores data.
- Has two pointers: left and right.
- Both pointers start as null (no children).
- Used to build binary trees by linking nodes.
- Example: node = {data:10, left:null, right:null}

Full Transcript

A binary tree node is created by assigning a data value and initializing two pointers, left and right, to null. This means the node has no children yet. The code example shows a class Node with a constructor setting these fields. The execution table traces the creation step by step, showing the node's data and pointers. Key points include understanding why pointers start as null and what the data field holds. The visual quiz tests knowledge of pointer values and node creation steps.

Practice

1. What does a typical binary tree node contain in its structure?

easy

A. A value and multiple child nodes in a list

B. A value, a left child node, and a right child node

C. Only a value and a pointer to the next node

D. Only a value without any child references

Solution

1. Step 1: Understand binary tree node components

   A binary tree node stores a value and has exactly two child references: left and right.

2. Step 2: Compare options with binary tree definition

   A value, a left child node, and a right child node correctly lists value, left child, and right child. Others describe linked list or general tree nodes.

3. Final Answer:

   A value, a left child node, and a right child node -> Option B

4. Quick Check:

   Binary tree node = value + left + right [OK]

Hint: Binary tree nodes always have left and right child pointers [OK]

Common Mistakes:

• Confusing binary tree node with linked list node
• Thinking nodes have multiple children like general trees
• Ignoring child pointers in node structure

2. Which of the following is the correct JavaScript syntax to define a binary tree node class?

easy

A. class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } }

B. function Node(value) { this.value = value; this.next = null; }

C. class Node { constructor(value) { this.value = value; this.children = []; } }

D. var Node = (value) => { this.value = value; this.left = null; this.right = null; }

Solution

1. Step 1: Identify correct class syntax for binary tree node

   class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } } uses ES6 class with constructor setting value, left, and right to null.

2. Step 2: Check other options for errors

   function Node(value) { this.value = value; this.next = null; } uses next pointer (linked list), C uses children array (general tree), D uses arrow function incorrectly for constructor.

3. Final Answer:

   class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } } -> Option A

4. Quick Check:

   Class with value, left, right = class Node { constructor(value) { this.value = value; this.left = null; this.right = null; } } [OK]

Hint: Use ES6 class with left and right set to null [OK]

Common Mistakes:

• Using 'next' instead of 'left' and 'right'
• Using arrow function as constructor
• Using children array instead of left/right

3. What will be the output of the following code?

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
console.log(root.left.value + ' -> ' + root.value + ' -> ' + root.right.value);

medium

A. "5 -> 10 -> 15"

B. "10 -> 5 -> 15"

C. "null -> 10 -> null"

D. "15 -> 10 -> 5"

Solution

1. Step 1: Trace node creation and assignments

   Root node value is 10, left child is 5, right child is 15.

2. Step 2: Evaluate console.log output

   Output concatenates left.value (5), root.value (10), right.value (15) with arrows.

3. Final Answer:

   "5 -> 10 -> 15" -> Option A

4. Quick Check:

   Left 5, root 10, right 15 = "5 -> 10 -> 15" [OK]

Hint: Print left, root, right values in order [OK]

Common Mistakes:

• Mixing order of values in output
• Confusing left and right child values
• Forgetting to access .value property

4. Identify the error in the following binary tree node code snippet:

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(20);
root.left = new Node(10);
root.right = new Node(30);
root.left.left = 15;
console.log(root.left.left.value);

medium

A. Missing constructor in Node class

B. Using console.log incorrectly

C. Assigning a number directly instead of a Node object to left.left

D. root.right is not assigned properly

Solution

1. Step 1: Analyze assignment to root.left.left

   root.left.left is assigned 15 (a number), not a Node instance.

2. Step 2: Understand why console.log fails

   Accessing .value on a number (15) causes error because 15 has no .value property.

3. Final Answer:

   Assigning a number directly instead of a Node object to left.left -> Option C

4. Quick Check:

   Child nodes must be Node objects, not raw values [OK]

Hint: Child nodes must be Node instances, not raw values [OK]

Common Mistakes:

• Assigning raw values instead of Node objects
• Ignoring error from accessing .value on number
• Assuming constructor is missing

5. Given the following binary tree node class, how would you write a function to count the total number of nodes in the tree starting from the root?

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

Choose the correct function implementation.

hard

A. function countNodes(node) { if (node === null) return null; return 1 + countNodes(node.left) + countNodes(node.right); }

B. function countNodes(node) { if (!node) return 1; return countNodes(node.left) + countNodes(node.right); }

C. function countNodes(node) { return node.value + countNodes(node.left) + countNodes(node.right); }

D. function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); }

Solution

1. Step 1: Understand base case for recursion

   If node is null, count is zero because no node exists there.

2. Step 2: Sum current node plus counts of left and right subtrees

   Return 1 (current node) plus recursive counts of left and right children.

3. Step 3: Check options for correct base case and sum

   function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); } correctly returns 0 for null and sums 1 + left + right counts.

4. Final Answer:

   function countNodes(node) { if (!node) return 0; return 1 + countNodes(node.left) + countNodes(node.right); } -> Option D

5. Quick Check:

   Count nodes = 1 + left count + right count [OK]

Hint: Return 0 for null, else 1 + left + right counts [OK]

Common Mistakes:

• Returning 1 for null node instead of 0
• Adding node.value instead of count 1
• Returning null instead of 0 in base case